Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.05-M HCl to 570. mL of each of the following solutions.

a) water  

before mixing = 7

After mixing = 1.45

b) 0.135 M C₂H₃O₂^1-

^{Before mixing = 8.95}

^{After mixing = ?}

c) 0.135 M HC₂H₃O₂

_{Before mixing = ?}

_{After mixing = ?}

d) a buffer solution that is 0.135 M in each C₂H₃O₂^1- and HC₂H₃O₂

_{Before mixing = ?}

_{After mixing = ?}

I've tried this so many times and I don't understand it. Is there a formula or a set of formulas for this ?? Can someone please just spell it out for me step by step and why. Please and thank you!

Answer 1

a) water
Before mixing: pH = 7.00
After mixing: [H+] is provided by HCl:
[H+] = 10*2.05/580 = 0.035

                     Hence: pH = log[H+] = 1.45
pH change = 7-1.45 = 5.55

b) 0.135 M C₂H₃O₂^1- :
Before mixing:
C₂H₃O₂^1- + H2O <==> CH3COOH + OH-

                     Kb = [CH3COOH]*[ OH-]/[ C₂H₃O₂^1-] = [CH3COOH]*[ OH-]*[H+]/[ C₂H₃O₂^1-][H+] = Kw/Ka
Notice here CH3COOH and OH- are generated in pairs, hence [CH3COOH] = [OH-], and
Kb = Kw/Ka = [CH3COOH]*[ OH-]/[ C₂H₃O₂^1-] = [OH-]^2/C
[OH-] = √(C*Kw/Ka)
[H+] = Kw/[OH-] = √( Kw.Ka/C)

                     log([H+]) = -log√( Kw.Ka/C)

                              = 0.5(logKw + logKa-logC)

                     pH = 0.5*(pKw +pKa +logC)

                              =0.5(14+4.76+log0.135)

                              = 8.95

After mixing:
The [H+] concentration provided by HCl in mixing is:
10*2.05/580=0.035M
The [C₂H₃O₂^1- ] concentration in mixing is:
0.135M*(570/580) = 0.133M
The reaction between H+ and C₂H₃O₂^1- forms CH3COOH and Cl-,

                     [CH3COOH] = 0.035M with [C₂H₃O₂^1-] = 0.133-0.035=0.098M left.

                     Kb = 5.75 X 10 ^-10= [CH3COOH]*[ OH-]/[ C₂H₃O₂^1-] = [OH-]*0.035/0.098
[OH-] = 1.61*10^-9

                     pOH = 8.79

                     pH = 14-8.79 =5.21

c) 0.135 M HC₂H₃O₂:

                     Before mixing:

                     CH3COOH + H2O = CH3COO^- + H3O⁺

                     Ka = [CH3COO^-] [H3O⁺]/ [CH3COOH]

                     =αC.αC /(1-α)C....[α is degree of dissociation]

                     = Cα² ...(As it is a weak acid, dissociation is very little, α is very small)

                     α = √ (Ka/C)

                     [H3O⁺] = αC = C* √ (Ka/C) = √ (Ka.C)

                     = √(1.73*10^-5*0.135

                     =1.53*10^-3

                     pH=2.81

After mixing:
Notice CH3COOH is a weak acid and therefore (1) there is no reaction between CH3COOH and HCl, and HCl provide more [H+] and CH3COOH dissociate much less.
The HCl or [H+] concentration after mixing is:
10*2.05/580=0.035M
pH = 1.45

d. a buffer solution that is 0.135 M in each C₂H₃O₂^1- and HC₂H₃O₂:

Before mixing: The [C₂H₃O₂^1- ] concentration or [CH3COOH] in mixing is:
0.135M*(570/580) = 0.133M

Apply Henderson equation for buffer solution,

pH = pKa + log [salt]/[acid]

= 4.76+ log0.133/0.133

=4.76

After mixing:

C₂H₃O₂^1- + H+ <==> CH3COOH

This concentration of salt decreases and concentration of acid increases.

[H+] is provided by HCl:
[H+] = 10*2.05/580 = 0.035

The [C₂H₃O₂^1- ] concentration or [CH3COOH] in mixing is:
0.135M*(570/580) = 0.133M

After reaction, [C₂H₃O₂^1- ] = 0.133-0.035=0.098 M

                         [CH3COOH] = 0.133+0.015 = 0.148 M

Apply Henderson equation for buffer solution,

pH = pKa + log [salt]/[acid]

= 4.76+ log0.098/0.148

=4.58